Question

What volume of 0.01000 M AgNO3; would be required to precipitate all the I - in 200.0 mL of a solution that contained 26.43 ppt KI?

Answer 1

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Answer 2

Given,

• 0.01000 M $AgNO_{3}$

• 26.43 ppm KI

To find,

The volume of $AgNO_{3}$ required to precipitate 200.0 mL of a solution that contains 26.43 ppm KI.

Solution,

We need to convert ppm to molarity.

The molar mass of KI = 166g

26.43ppm = 26.43 grams of KI in $10^{6}$ gm of water, i.e 1000 lt of water.

Molarity = $\frac{26.43}{166*1000} = 0.000159M$

To find the volume of $AgNO_{3}$ required, we need to establish a relation between the both.

The reaction taking place will be,

$AgNO_{3}$ $+ KI ---> AgI + KNO_{3}$

Moles = molarity*volume.

As the valency of Ag and K is 1, same moles of both the compounds will be required to precipitate AgI completely without leaving any unreacted compound.

 Moles of $AgNO_{3}$ = Moles of $KI$

     $0.01*v = 200*0.000159\\v = 3.18ml$

Therefore, $3.18ml$ of 0.01000 M AgNO3 is required to precipitate all the I - in 200.0 mL of a solution that contained 26.43 ppm KI.